What is the integral from 1 to 2 of (10x^3+6/(x^2)) ?

The integral from 1 to 2 of (10x^3+6/(x^2)) is 81/2

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integral from 1 to 2 of (10x^3+6/(x^2))

\int_{1}^{2} (10 x^{3} + \frac{6}{x ^{2}}) d x

\frac{81}{2}

Solution steps

\int_{1}^{2} 10 x^{3} + \frac{6}{x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} 10 x^{3} d x + \int_{1}^{2} \frac{6}{x ^{2}} d x

\int_{1}^{2} 10 x^{3} d x = \frac{75}{2}

\int_{1}^{2} \frac{6}{x ^{2}} d x = 3

= \frac{75}{2} + 3

Simplify

\frac{75}{2} + 3 : \frac{81}{2}

= \frac{81}{2}

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What is the integral from 1 to 2 of (10x^3+6/(x^2)) ?
The integral from 1 to 2 of (10x^3+6/(x^2)) is 81/2